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2 years ago

why did mendeleev leave blank spaces on his periodic table? did later discoveries justify his predictions?

Chemistry

1 answer:

schepotkina [342]2 years ago

7 0

Answer:

mendeleev left a space

Explanation:

so the periodic table can be organize

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An electrochemical cell has the following standard cell notation: Al(s) | Al3+(aq) || Mg2+(aq) | Mg(s)

Soloha48 [4]

Answer:

a. Al(s) ⇄ Al⁺³(aq) + 3e⁻ (oxidation)

Mg²⁺(aq) + 2e⁻ ⇄ Mg(s) (reduction)

b. ΔE° = + 0.715 V

c. It's an electrolytic cell, because it's a nonspontaneous reaction.

d. 2Al(s) + 3Mg²⁺(aq) ⇄ 2Al⁺³(aq) + 3Mg(s)

Explanation:

a. By the notation given, first is represented the oxidation reaction and then the reduction reaction, so they are:

Al(s) ⇄ Al⁺³(aq) + 3e⁻ (oxidation)

Mg²⁺(aq) + 2e⁻ ⇄ Mg(s) (reduction)

b. The standard potential of the cell (ΔE°) is the reduction potential of the oxidation less the reduction potential of the reduction. The reduction potentials are:

Al(s) = -1.66 V

Mg(s) = -2.375 V

ΔE° = -1.66 - (-2.375)

ΔE° = + 0.715 V

c. It's an electrolytic cell.

A galvanic cell is spontaneous, so the cathode (reduction) has a higher E° than the cathode (oxidation). In this case, the oxidation reaction has a higher E°, so the reaction is nonspontaneous and it's necessary an external force to it happen, so it's an electrolytic cell.

d. 2Al(s) + 3Mg²⁺(aq) ⇄ 2Al⁺³(aq) + 3Mg(s)

The number of electrons must be the same, so the oxidation reaction is multiplied by 2, and the reduction reaction by 3.

5 0

3 years ago

Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 9.32 g of ethane is

Pie

Answer:

There will remain 8.06 grams of ethane

Explanation:

Step 1: Data given

Mass of ethane = 9.32 grams

Mass of oxygen = 12.0 grams

Molar mass ethane = 30.07 g/mol

Molar mass oxygen = 32.00 g/mol

Step 2: The balanced equation

2 C2H6 + 7 O2 → 4 CO2 + 6 H2O

Step 3: Calculate moles ethane

Moles ethane = mass ethane / molar mass ethane

Moles ethane = 9.32 grams / 30.07 g/mol

Moles ethane = 0.3099 moles

Step 4: Calculate moles oxygen

Moles oxygen = 12.0 grams / 32.0 g/mol

Moles oxygen = 0.375 moles

Step 5: Calculate the limiting reactant

For 2 moles ethane we need 7 moles O2 to produce 4 moles CO2 and 6 moles H2O

O2 is the limiting reactant. It will completely be consumed ( 0.375 moles)

Ethane is in excess. There will react 2/7 * 0.375 = 0.107 moles

There will remain 0.375 - 0.107 = 0.268 moles

Step 6: Calculate mass ethane

Mass ethane = moles ethane * molar mass ethane

Mass ethane = 0.268 moles * 30.07 g/mol

Mass ethane = 8.06 grams

There will remain 8.06 grams of ethane

7 0

3 years ago

Why do particles settle down in suspension

fomenos

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6 0

3 years ago

Balance the following Chemical Equation:<br> NaBr +CaCl2-> NaCl+ CaBr2

frozen [14]

First write all of the compounds/atoms in either side then fill in existing values and balance

Na- 1
Br- 1
Ca- 1
Cl- 2

Na- 1
Cl- 1
Ca-1
Br-2

Balance to get

2NaBr+CaCl2=2NaCl+CaBr2

7 0

3 years ago

How many hydrogen atoms are in 105 g of water?

mel-nik [20]

Moles of water atoms = mass/molecular weight = 105/18 = 5.83 mol. Number of moles of hydrogen in water = 2 x moles of water = 11.66. Number of H atoms in water = moles of hydrogen x 6.02 x 10^23 = 7.019 x 10^24 ~ 7.02 x 10^24 atoms. Hope this helps.

3 0

3 years ago